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    Find an element of order 3 in the group GL(2, R) where R is the set of all real numbers.

    I think that such an element does not exist. Am I correct? We want a "cube root" of the identity matrix but it can't be the identity matrix itself because the identity matrix has order 1. The other two "cube roots" of the identity matrix do not have real entries and hence are not in this group.
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    You are not correct. Think rotations.
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    (Original post by generalebriety)
    You are not correct. Think rotations.
    I see. So we could have a rotation of 2pi/3 so that when we cube this matrix and do the rotation 3 times we get a rotation of 2pi, which keeps everything the same.

    But aren't the three "cube roots" of the identity matrix the only three matrices that cube to give us the identity matrix? With the above, this is clearly not true. But why not?
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    (Original post by gangsta316)
    I see. So we could have a rotation of 2pi/3 so that when we cube this matrix and do the rotation 3 times we get a rotation of 2pi, which keeps everything the same.

    But aren't the three "cube roots" of the identity matrix the only three matrices that cube to give us the identity matrix? With the above, this is clearly not true. But why not?
    I'm not sure why you think the identity matrix only has three cube roots. It has infinitely many.
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    (Original post by generalebriety)
    I'm not sure why you think the identity matrix only has three cube roots. It has infinitely many.
    Because 1 has three cube roots and (because the identity matrix is diagonal) we cube root the identity matrix by cube rooting its entries.
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    (Original post by gangsta316)
    Because 1 has three cube roots and (because the identity matrix is diagonal) we cube root the identity matrix by cube rooting its entries.
    I don't know where you've got that idea from. If you want to raise some diagonal matrix D to the power of a positive integer, then yes, just raise each of its entries to that power; so if D^3 = I, we can find three obvious diagonal cube roots of I that way. The converse isn't true - there's no reason you can't have a non-diagonal matrix whose cube is diagonal. In fact, you've found at least one yourself - that rotation matrix R (and another, its inverse). And, in fact, P-1RP and Q-1DQ are cube roots of the identity for any invertible matrices P, Q. (Try it.)
 
 
 
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